3.5 Perpendicular bisectors

Introduction

A new bus stop should be the same distance from two schools; a border at sea should be fair to two countries; a phone should connect to whichever of two masts is closer. Each of these problems asks for the points that are the same distance from two fixed points, and those points always form one straight line, the perpendicular bisector of the segment joining them.

The perpendicular bisector is the building block of the Voronoi diagrams in subtopic 3.6, where every boundary between two regions is part of one. It is also the quickest route to the centre of a circle through three known points, which surveyors and archaeologists use to recover the centre of a circular feature from a few points on its edge.

The tools from straight lines

Finding a perpendicular bisector combines four short results from earlier work, and every bisector calculation uses all four.

The midpoint of a line segment is the point halfway between its ends, found by averaging the coordinates. The gradient measures the steepness of the segment, and two lines are perpendicular when the product of their gradients is $-1$, so the gradient of a perpendicular line is the negative reciprocal of the original. A line is fixed by one point and its gradient, which point-gradient form turns straight into an equation.

What a perpendicular bisector is

To bisect something is to cut it into two equal parts. The word is used for angles as well as segments.

A segment has infinitely many bisectors: every line through its midpoint cuts it in half. Exactly one of them crosses it at a right angle.

The diagram of a perpendicular bisector shows both properties at once: the bisector meets $[AB]$ at its midpoint $M$ at a right angle, and a point $P$ anywhere on it is the same distance from $A$ as from $B$.

Equidistance

Equidistant points

The second half of the definition, equal distance from the two endpoints, is the property that applications use.

The collection of all points that satisfy a condition is described with its own word.

Why every point on the bisector is equidistant

Let $M$ be the midpoint of $[AB]$ and $P$ any point on the perpendicular bisector. The triangles $PMA$ and $PMB$ each have a right angle at $M$, share the side $PM$, and have equal sides $AM = MB$. Pythagoras' theorem in each triangle gives

so $PA = PB$. The two triangles are mirror images in the bisector, which is the same fact seen as a symmetry: reflecting $A$ in the perpendicular bisector gives $B$.

Why every equidistant point is on the bisector

The converse also holds: no point off the bisector is equidistant from $A$ and $B$. The distance formula proves both directions together. A point $(x, y)$ is equidistant from $A(x_1, y_1)$ and $B(x_2, y_2)$ exactly when the squared distances are equal.

The $x^2$ and $y^2$ terms cancel, so the equidistant points satisfy a linear equation: they form a straight line. Its gradient is $-\frac{x_2 - x_1}{y_2 - y_1}$, the negative reciprocal of the gradient of $[AB]$, so it is perpendicular to $[AB]$; and substituting the midpoint makes both sides equal $x_2^2 - x_1^2 + y_2^2 - y_1^2$, so it passes through the midpoint. The line of equidistant points is therefore exactly the perpendicular bisector.

The equation of a perpendicular bisector from two points

The method

Given $A(x_1, y_1)$ and $B(x_2, y_2)$, four steps give the equation.

1. Find the midpoint $M$ of $[AB]$. The bisector passes through it.

2. Find the gradient $m$ of $[AB]$.

3. Find the perpendicular gradient $-\frac{1}{m}$. This is the bisector's gradient.

4. Substitute $M$ and $-\frac{1}{m}$ into $y - y_1 = m(x - x_1)$

Steps 1 and 4 make the line pass through the midpoint, and steps 2 and 3 make it perpendicular, so together they meet both halves of the definition.

The check at the end uses the equidistant property, and it catches the usual errors: a line through an endpoint, a line with the wrong gradient and most arithmetic slips all give a point that is not equidistant.

Integer coefficients

An equation is often wanted in the form $ax + by + d = 0$ where $a, b, d \in \mathbb{Z}$. Multiplying through by the lowest common denominator of the fractions clears them, and moving every term to one side gives the general form.

The distance-formula method

The equidistance argument gives a second method that needs no gradients at all. Setting the squared distances from $(x, y)$ to the two points equal and expanding gives the bisector directly in general form.

The gradient method shows each property of the bisector as a separate, visible step (midpoint, gradient, negative reciprocal, substitution), which makes errors easy to locate. The distance-formula method is a useful independent check and works well when the answer is wanted in general form.

Horizontal and vertical segments

When the two points have the same $y$-coordinate, the segment is horizontal, its gradient is 0, and the negative reciprocal $-\frac{1}{0}$ does not exist. The bisector is vertical, so its equation is $x = k$, where $k$ is the $x$-coordinate of the midpoint. In the same way, a vertical segment (same $x$-coordinate) has a horizontal bisector $y = k$.

Checking with the GDC

A GDC gives a visual check. Plot the two points (as a short list in the statistics editor, or by eye) and graph the bisector; it should cross the segment halfway and at a right angle. Right angles look like right angles only when both axes have the same scale, so use a window with equal scales, such as the zoom menu's square option (TI-84 Plus CE: ZOOM → ZSquare). On paper, the check is the equidistance of one point on the line, as in the first worked example.

From a segment's equation and its midpoint

Sometimes the endpoints are not known. Instead, the information available is the equation of the line containing the segment and the coordinates of its midpoint. The method is the same, except that the gradient is read from the equation rather than calculated from two points.

A useful shortcut for a line in general form: $ax + by + d = 0$ has gradient $-\frac{a}{b}$, and any line perpendicular to it can be written $bx - ay + k = 0$ (swap the coefficients and change one sign). For $2x - 3y + 6 = 0$, perpendicular lines are $3x + 2y + k = 0$, and $M(3, 4)$ gives $k = -17$.

Where a bisector meets another line

Many problems need a point that is equidistant from two places and lies on a given road, river or pipeline. Such a point lies on two lines at once, the bisector and the given line, so it is their intersection. Finding it means solving two linear equations together, a system of linear equations: by substitution when both are in the form $y = mx + c$, or with the GDC's simultaneous equation solver (TI-84 Plus CE: APPS → PlySmlt2 → SIMULTANEOUS EQN SOLVER; Casio fx-CG50: MENU → Equation → SIMUL) or the intersect tool on a graph.

The graph of the villages and the road shows the construction: the bisector crosses the segment $[AB]$ at its midpoint at a right angle, and meets the road at the only point on the road that is fair to both villages.

The reverse problem: finding an endpoint

If one endpoint $B$ and the perpendicular bisector are known, the other endpoint $A$ can be recovered. $A$ is the mirror image of $B$ in the bisector, so three steps find it.

1. The segment $[AB]$ is perpendicular to the bisector, so find the line through $B$ with the negative reciprocal of the bisector's gradient.

2. Intersect that line with the bisector. The intersection is the midpoint $M$.

3. $M$ is halfway from $B$ to $A$, so $A = (2x_M - x_B,\ 2y_M - y_B)$.

Points equidistant from three points

The circumcentre

A point equidistant from $A$, $B$ and $C$ must lie on the perpendicular bisector of $[AB]$ (equidistant from $A$ and $B$) and on the perpendicular bisector of $[BC]$ (equidistant from $B$ and $C$). Two non-parallel lines meet at exactly one point, so if the three points do not lie on one straight line there is exactly one such point.

The third bisector needs no calculation. A point with $OA = OB$ and $OB = OC$ also has $OA = OC$, so it lies on the bisector of $[AC]$ too: all three bisectors pass through the same point. Because the circumcentre is the same distance from all three vertices, a circle centred there passes through all three.

Where the circumcentre lies depends on the shape of the triangle. For an acute triangle it is inside; for a right-angled triangle it is the midpoint of the hypotenuse, so the hypotenuse is a diameter; for an obtuse triangle it is outside. If the three points lie on a straight line, the bisectors are parallel and no point is equidistant from all three.

Finding the centre of a circle from three points

Three points on the edge of any circle, such as the stones of a buried circular wall found by archaeologists, determine the circle. Its centre is the circumcentre of the triangle they form.

The graph of the post holes shows the two bisectors meeting at the centre and the circle of radius 5 m through all three holes.

Equidistance lines and fair boundaries

When two countries face each other across a narrow sea, their territorial seas (which extend up to 12 nautical miles from the coast) can overlap. International law then sets the default boundary as the median line, every point of which is equidistant from the nearest points on the two countries' baselines (lines that mostly follow the low-water coastline), unless the countries agree otherwise or historic title or special circumstances call for a different line.

Source: United Nations Convention on the Law of the Sea, Article 15 (UN text, accessed 2026); Wikipedia, "Territorial waters", accessed 2026.

For two points the equidistance line is a single perpendicular bisector. A real coastline has many points, so the median line is built from pieces of bisectors: along each stretch it follows the bisector of the two coastal points that are currently nearest, and it turns where the nearest pair changes. The result has the same shape as the edges of a Voronoi diagram, with the coastal points as sites.

The same reasoning applies to two identical phone masts. If signal strength depends only on distance, a phone on the perpendicular bisector of the two masts receives equal signals, and on either side it connects to the nearer mast. Adding a third mast and more bisectors divides a whole region into the cells of a Voronoi diagram.

Ruler-and-compass construction

Long before coordinates, the perpendicular bisector was drawn with a straight edge and a pair of compasses. The construction is not needed when coordinates are given, but it shows the equidistant property directly.

To bisect $[AB]$:

1. Open the compasses to a radius $r$ more than half the length of $[AB]$.

2. With the point on $A$, draw arcs above and below the segment.

3. With the same radius and the point on $B$, draw arcs that cross the first two at $X$ and $Y$.

4. Draw the straight line through $X$ and $Y$. It is the perpendicular bisector, and it crosses $[AB]$ at its midpoint $M$.

The diagram of the construction shows why it works: $X$ is a distance $r$ from both $A$ and $B$, so it is equidistant from them, and so is $Y$. Both points lie on the perpendicular bisector, and two points fix a line.

If the radius is less than half of $AB$, the arcs do not meet, which is why step 1 needs a radius greater than half the segment. The four points $A$, $X$, $B$, $Y$ form a rhombus with all sides $r$, and the diagonals of a rhombus bisect each other at right angles, which is a second explanation of the result.

Exam focus

Perpendicular bisectors appear on Paper 1 as questions of 4–7 marks and inside longer Voronoi diagram questions on Paper 2 (and, at HL, Paper 3). The command terms are find (show the midpoint, the gradient, the perpendicular gradient and the substitution), write down (a bisector such as $x = 14$ for a horizontal segment, needing no working), determine (a point equidistant from two places on a given line) and "give your answer in the form …" ($y = mx + c$, or $ax + by + d = 0$ with $a, b, d \in \mathbb{Z}$). A typical markscheme for a 4–5 mark bisector awards (A1) for the gradient of the segment, (A1) for the midpoint, (M1) for using the negative reciprocal, (M1) for substituting the midpoint into a line equation, and A1 for the final equation in the requested form. Midpoints must be written as coordinate pairs, in brackets.

In part (d) the segment $[AC]$ is horizontal, so no gradient is needed: the bisector is the vertical line through the midpoint $(5, 4)$. Part (e) uses "hence", so the expected method intersects the bisectors already found rather than starting again.

Practice questions

1. Find the midpoint and the gradient of the segment joining $(-2, 1)$ and $(6, 5)$. [3]

2. Find the equation of the perpendicular bisector of the segment joining $(-2, 1)$ and $(6, 5)$, in the form $y = mx + c$. [3]

3. Write down the equation of the perpendicular bisector of the segment joining $(5, -3)$ and $(5, 11)$. [1]

4. Find the perpendicular bisector of $S(-1, 4)$ and $T(5, -6)$ in the form $ax + by + d = 0$

5. A segment lies on the line $y = 4x - 7$ and has midpoint $(2, 1)$. Find the equation of its perpendicular bisector. [3]

6. Two wells are at $W_1(0, 6)$ and $W_2(8, 2)$ (in km). A pipeline follows $y = -x + 11$

7. The perpendicular bisector of $[CD]$ is $y = -x + 4$, and $D$ is $(5, 3)$. Find the coordinates of

8. Find the coordinates of the point equidistant from $E(0, 0)$, $F(8, 0)$ and $G(0, 6)$, and explain why it is the midpoint of $[FG]$

9. Show that the point $(1, 7)$ lies on the perpendicular bisector of $A(-2, 3)$ and $B(5, 4)$. [3]

10. Explain why three points on a straight line have no point that is equidistant from all three. [2]

Summary

• A bisector cuts a segment or angle into two equal parts; the perpendicular bisector of $[AB]$ passes through its midpoint at right angles.

• Every point on the perpendicular bisector is equidistant from $A$ and $B$, and every equidistant point is on it: it is the locus of points equidistant from $A$ and $B$.

• Method: midpoint, gradient of $[AB]$, negative reciprocal, then $y - y_1 = m(x - x_1)$ through the midpoint.

• Horizontal segments have vertical bisectors $x = k$; vertical segments have horizontal bisectors $y = k$.

• Setting $PA^2 = PB^2$ for a general point gives the bisector directly in general form.

• Given a segment's equation and midpoint, take the gradient from the equation and proceed as before.

• The point on a given line equidistant from two points is the intersection of that line with their bisector.

• Two bisectors of a triangle's sides meet at the circumcentre, the centre of the circle through all three vertices.

Self-check

• To do:

  I can state the definition of a perpendicular bisector and of a locus.

• To do:

  I can explain, with Pythagoras' theorem or the distance formula, why points on the bisector are equidistant from the endpoints.

• To do:

  I can find the midpoint of a segment and write it as coordinates in brackets.

• To do:

  I can find the perpendicular gradient and check it by multiplying to get $-1$.

• To do:

  I can find the equation of a perpendicular bisector in the form $y = mx + c$.

• To do:

  I can rearrange it into $ax + by + d = 0$ with integer coefficients.

• To do:

  I can write down the bisector of a horizontal or vertical segment.

• To do:

  I can find a bisector from a segment's equation and midpoint.

• To do:

  I can find where a bisector meets another line, by substitution or with my GDC.

• To do:

  I can find a missing endpoint from the other endpoint and the bisector.

• To do:

  I can find the circumcentre of three points and the radius of the circle through them.

• To do:

  I can check an answer by showing that a point on my line is equidistant from both points.

• To do:

  I can explain why the edges of a Voronoi diagram are parts of perpendicular bisectors.

$(5, 5)$ A1

[2 marks]

Gradient of $[AB]$:

The perpendicular gradient is $-\frac{3}{4}$. Through $M(6, 2)$:

The perpendicular bisector is $y = -0.75x + 6.5$.

Check with a point on the line, such as $(2, 5)$: its distance from $A$ is $\sqrt{1^2 + 7^2} = \sqrt{50}$ and from $B$ is $\sqrt{7^2 + 1^2} = \sqrt{50}$. ✓

Gradient of $[PQ]$: $\frac{5 - 2}{6 - 1} = \frac{3}{5}$, so the perpendicular gradient is $-\frac{5}{3}$.

The perpendicular bisector is $5x + 3y - 28 = 0$. Any non-zero integer multiple, such as $10x + 6y - 56 = 0$, is also correct.

Dividing by 2 gives $5x + 3y - 28 = 0$, the same line as the gradient method.

(b) $[EF]$ is vertical (both $x$-coordinates are 4). Its midpoint is $\left(4, \frac{-1 + 9}{2}\right) = (4, 4)$, and the bisector is the horizontal line through it: $y = 4$.

The perpendicular gradient is $-\frac{3}{2}$. Through $M(3, 4)$:

The perpendicular bisector is $3x + 2y - 17 = 0$.

(b) The centre lies on both lines:

The health centre is at $(8, 8)$.

(c) $\sqrt{(8 - 1)^2 + (8 - 9)^2} = \sqrt{50}$ and $\sqrt{(8 - 7)^2 + (8 - 1)^2} = \sqrt{50}$, so each village is $\sqrt{50} = 7.07\text{ km}$ (3 s.f.) away. The equal distances confirm the answer.

Intersect with the bisector to find the midpoint:

So $M = (3, 5)$. Since $M$ is the midpoint of $[AB]$,

Check: the gradient of $[AB]$ is $\frac{2 - 8}{9 - (-3)} = -\frac{1}{2}$, perpendicular to 2 ✓, and the midpoint is $(3, 5)$, which lies on $y = 2x - 1$ ✓.

(b) Substitute $y = 7x - 36$ into $x + 7y - 48 = 0$:

With the GDC's simultaneous equation solver, enter $7x - y = 36$ and $x + 7y = 48$. The centre is $(6, 6)$.

(c) Radius $= \sqrt{(6 - 2)^2 + (6 - 3)^2} = \sqrt{25} = 5\text{ m}$. Check: $\sqrt{3^2 + 4^2} = 5$ to $Q$ and $\sqrt{4^2 + 3^2} = 5$ to $R$ ✓.

For the ship, compare distances: to $K$, $\sqrt{10^2 + 4^2} = \sqrt{116}$; to $L$, $\sqrt{6^2 + 8^2} = \sqrt{100}$. The ship is closer to $L$, so it is in $L$'s waters. As a check, $4(10) + 3(4) - 50 = 2 > 0$, the same sign as for $L$ ($64 + 36 - 50 = 50 > 0$), while $K$ gives $-50 < 0$: the ship is on $L$'s side of the line.

Note: award A0 if the brackets are omitted.

(b) gradient of $[AB]$ $= \frac{8 - 4}{10 - 2} = \frac{1}{2}$ (A1)

gradient of bisector $= -2$ (M1)

$y - 6 = -2(x - 6)$ (M1)

$2x + y - 18 = 0$ A1

(c) $-2x + 18 = x - 3$ (or a GDC intersection of the two lines) (M1)

$(7, 4)$ A1

(d) $x = 5$ A1

(e) substituting $x = 5$ into $2x + y - 18 = 0$ (M1)

$(5, 8)$ A1

distance $= \sqrt{(5 - 2)^2 + (8 - 4)^2}$ (M1)

$= 5\text{ km}$ A1

Total [13]